Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the compound \((\mathrm{X})\) and the condition (Y) her \(\mathrm{PbS} \underset{\text { in air }}{\stackrel{\text { Heat }}{\longrightarrow}(\mathrm{X})+\mathrm{PbS}} \stackrel{(\mathrm{Y})}{\longrightarrow} \mathrm{Pb}+\mathrm{SO}_{2}\) (a) \(\mathrm{X}=\mathrm{PbSO}_{3}, \mathrm{Y}=\) low temperature (b) \(\mathrm{X}=\mathrm{PbO}\) or \(\mathrm{PbSO}_{4}, \mathrm{Y}=\) high temperature (air supply is cut off) (c) \(\mathrm{X}=\mathrm{Pb}_{3} \mathrm{O}_{4}, \mathrm{Y}=\) high temperature (d) \(\mathrm{X}=\mathrm{PbSO}_{4}, \mathrm{Y}=\mathrm{N}_{2}\)

Step-by-Step Solution

Answer

Option (b): \( \text{X} = \text{PbO or PbSO}_4, \text{Y} = \) high temperature (air supply is cut off).

Step 1: Identify the first reaction

The reaction \( \text{PbS} \underset{\text{ in air }}{\stackrel{\text{Heat}}{\longrightarrow}} (\mathrm{X}) + \text{PbS} \)suggests the formation of a lead compound \((X)\) by heating lead sulfide \(\text{PbS}\) in the presence of air. Typically, \(\text{PbS}\) reacts with oxygen in the air to form lead oxide \(\text{PbO}\) or lead sulfate \(\text{PbSO}_4\).

Step 2: Determine suitable compound \(X\)

In the "in air" heating scenario, \(\text{PbS}\) can oxidize to form either \(\text{PbO}\) or \(\text{PbSO}_4\). Later reactions provide clues by looking at the resulting reactions and whether oxygen or sulfur mainly contributes equilibrated products under specific temperatures and conditions.

Step 3: Analyze the second reaction and condition \(Y\)

The second reaction is \( (\mathrm{X}) + \text{PbS} \stackrel{(\mathrm{Y})}{\longrightarrow} \text{Pb} + \text{SO}_2 \). If \((X)\) is \(\text{PbO}\), it can further react with \(\text{PbS}\) upon "cut off air supply" high temperatures, allowing lead \((\text{Pb})\) and \(\text{SO}_2\) to form. Thus, high temperature with cut off air fits for \(Y\).

Step 4: Identify the correct option

Given the feasible reactions and conditions outlined in Steps 1-3, the reaction set matches (b) \( \text{X} = \text{PbO or PbSO}_4, \text{Y} = \) high temperature (air supply is cut off), since \(\text{PbO}\) can decompose into lead when preventing further oxidation.

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, are essential in understanding chemical behavior. In a redox reaction, there is a transfer of electrons between two substances. One substance will lose electrons and undergo oxidation. The other substance will gain those electrons and undergo reduction.

In the context of the given problem, we observe redox reactions involving lead sulfide (\(\text{PbS}\)) and the formation of compounds like lead oxide (\(\text{PbO}\)). The lead in \(\text{PbS}\) can be oxidized when heated in air, reacting with oxygen (\(\text{O}_2\)) to form lead oxide and possibly lead sulfate (\(\text{PbSO}_4\)) depending on the conditions.

These reactions highlight the essential role oxidation and reduction play in transforming reactive materials into different products. To identify whether a substance undergoes oxidation or reduction, look at the change in oxidation number. An increase in oxidation number indicates oxidation, while a decrease indicates reduction. In redox reactions involving lead compounds, understanding these changes allows for predicting the outcomes of reactions.

Lead Compounds

Lead compounds, like \(\text{PbO}\) and \(\text{PbSO}_4\), are formed through specific chemical environments and reactions. Lead oxide (\(\text{PbO}\)) is common and forms readily when lead sulfide (\(\text{PbS}\)) reacts with oxygen from the air during heating processes.

Lead, being a heavy metal, forms various compounds with significant commercial relevance and historical use, such as in paints and batteries.

• \(\text{PbO}\) - Generally appears yellow or red, used in making glass and ceramics.
• \(\text{PbSO}_4\) - White or colorless, used in batteries and pigments.
• Other forms like \(\text{Pb}_3\text{O}_4\) are less common.

Understanding how to manipulate reaction conditions to produce specific lead compounds is crucial for applications in industry. Safety is paramount with lead compounds due to toxicity and environmental concerns. Proper handling and disposal are necessary to mitigate health risks.

Chemical Reaction Mechanisms

Chemical reaction mechanisms provide insight into the step-by-step processes that occur during a reaction. They help chemists understand not just the reactants and products, but the process and pathway the reaction takes to go from one to another.

For the Ncert exercise, the mechanism involves heating \(\text{PbS}\) in air where it undergoes oxidation to form lead oxide \(\text{PbO}\). When the supply of air is cut, and the reaction is carried out at high temperatures, \(\text{PbO}\) reacts with remaining \(\text{PbS}\) to produce metallic lead \(\text{Pb}\) and sulfur dioxide \(\text{SO}_2\).

• Step 1: \(\text{PbS (s) + O}_2\text{ (g)} \rightarrow \text{PbO (s)}\)
• Step 2: \(\text{PbO (s) + PbS (s)} \rightarrow 2\text{Pb (s) + SO}_2\text{ (g)}\)

This systematic approach helps in predicting reactions' progress, understanding conditions required, and ensuring desired products form. Mechanisms are foundational in designing and controlling chemical reactions, from industrial applications to everyday chemistry.